 # Composition Of Functions & Inverse Of A Function

## Composite Functions

When two functions are combined in such a way that the output of one function becomes the input to another function, then this is referred to as composite function. Consider three sets X, Y and Z and let f : X → Y  and g: Y → Z.

According to this, under map f, an element  x ∈ X is mapped to an element y = f(x) ∈ Y which in turn is mapped by g to an element z ∈ Z  in such a manner that z = g(y) = g[(f(x)] .

This mapping comprising of mappings f and g is known as composition of mappings. It is denoted by gof . Therefore, we are mapping onto .

The composite function is denoted as:

$$~~~~~~~~~$$ ( gof)(x) = g(f (X) )

Similarly, (fog) (x) = f (g(x))

So, to find (gof) (x), take f(x)  as argument for the function g.

Let us try to solve some questions based on composite functions.

 Let’s Work Out: Example: Given the function f(x) = 3x + 5  and g(x) = $$2x^3$$  .Find ( gof)(x) and ( fog)(x). Solution:  We know, (gof) (x) = g(f(x)) = g (3x+5) = $$2(3x+5)^3$$ Using Binomial Expansion, we have $$(gof) (x) = 2 \left [ (3x)^{3} + 3.(3x)^{2}.5 + 3.(3x)(5)^{2}+(5)^3 \right ]$$ $$\Rightarrow (gof) (x) = 2 \left [ 27x^{3} + 135x^{2} + 225 x + 125 \right]$$ $$\Rightarrow (gof) (x) = 54x^{3} + 270x^{2} + 450 x + 250$$ Now, (fog)(x) = f $$\left( g(x) \right)$$ = f ( $$2x^3$$) = $$3 (2x^{3}) + 5$$ $$\Rightarrow (fog)(x) = 6x^{3} + 5$$ Example: Let f(x) = $$x^2$$ and g(x) = $$\sqrt{1 – x^2}$$ .Find (gof)(x) and ( fog)(x) . Solution: (gof)(x) = g( f(x)) = g($$x^2$$) = $$\sqrt{1 – (x^2)^2}$$ = $$\sqrt{1 – x^4}$$ (fog) (x) = f(g(x)) = f( $$\sqrt{1 – x^2}$$) = ( $$\sqrt{1 – x^2})^2$$ = $$1 – x^2$$

## Inverse of a Function

Let f:X → Y. Now, let f represent a one to one function and y be any element of Y , there exists a unique element x ∈ X  such that y = f(x).Then the map

$$~~~~~~~~~~$$ $$f^{-1}:f[X] \rightarrow X$$

which associates to each element is called as the inverse map of f.

The function f(x) = $$x^5$$ and g(x) = $$x^{\frac 12}$$   have the following property:

$$f\left( g(x) \right)$$ = $$f \left( x^{\frac 15} \right)$$ = $$(x^{\frac 15} )^5$$ = x

$$g\left( f(x) \right)$$ = $$g \left( x^{5} \right)$$ = $$(x^{5} )^{\frac 15}$$ = x

Thus, if two functions f and g  satisfy $$f \left( g(x) \right)$$ = x for every x in domain of f , then in such a situation we can say that the function f is the inverse of g  and g  is the inverse of f .

For finding the inverse of a function,we write down the function y as a function of x  i.e. y = f(x)   and then solve for x  as a function of y.

To have a better insight on the topic let us go through some examples.

 Let’s Work Out: Example: If f(x) = $$x^2$$, g(x) = $$\frac{x}{3}$$   and h(x) = 3x+2 . Find out fohog(x). Solution:  $$h\left( g(x) \right)$$ = 3 $$\left ( \frac x3 \right)$$ + 2 = x + 2 fohog(x) = f $$\left[ h\left(g(x)\right)\right]$$ = $$( x+ 2)^2$$ This is the required solution. Example: Example 2: Find the inverse of the function f(x) = $$x^3$$ , x ∈ R. Solution: The given function f(x) = $$x^3$$   is a one to one and onto function defined in the range → R  . Therefore, we can find the inverse of this function. To find the inverse, we need to write down this function as $$~~~~~~~~~~~~~$$ y = $$x^3$$ In the above equation,y  is an arbitrary element from the range of f. If we solve for x from the above equation, we will get: $$~~~~~~~~~~~~~$$ x = $$y^{\frac 13}$$ This gives a function g:Y →X. This new function g can be defined as $$~~~~~~~~~~~~~$$ g(y) = $$y^{\frac 13}$$ This function g is the inverse of the function f since it’s domain is same as the range of the function f.Since g(y) = $$y ^{\frac 12}$$ , representing the independent variable with x , we get g(x) = $$x^{\frac 13}$$ = $$f^{-1}(x)$$..